## Patterns | 11, 1111, 111111, …

Note the pattern:

11   =   (1 * 9)   +   2
1111   =   (123 * 9)   +   4
111111   =   (12345 * 9)   +   6
11111111   =   (1234567 * 9)   +   8
1111111111   =   (123456789 * 9)   +   10
111111111111   =   (12345679011 * 9)   +   12
11111111111111   =   (1234567901233 * 9)   +   14
1111111111111111   =   (123456790123455 * 9)   +   16
111111111111111111   =   (12345679012345677 * 9)   +   18
11111111111111111111   =   (1234567901234567899 * 9)   +   20

And so on.

11,   1111,   111111, …   have all an even number of 1s,
Each of these numbers are expressed as   addend(1) + addend(2),   where addend(2) represents the number of 1s

If   111…111   has n (= 2k) times 1s
Can   111…111   be expressed as   addend(1)   +   n ?

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11,   1111,   111111, …   have all an even number of 1s,

11   =   2   +   3^2
1111   =   22   +   33^2
111111   =   222   +   333^2
11111111   =   2222   +   3333^2
1111111111   =   22222   +   33333^2

Prove that

If   111….111   has n (= 2k) times 1s, then   22…22   has k times 2s and   33…33   has k times 3s